A hotel chain asks its patrons to complete a survey on...

A hotel chain asks its patrons to complete a survey on "satisfaction." So, naturally, they are hoping that their score is significantly different from that of their competitor's rating of 5.6. 1. Hotel Competition: Which test should they run to determine if the price is different? Group of answer choices One sample t-test Paired sample t-test Chi-Square Independent t-test 2. Run the appropriate test. What p-value would your report? Group of answer choices .15582 1.129 .260 .17600 3. Hotel Competition: Should the hotel be happy? Is there a significant difference between their ratings for satisfaction and that of their competitors? Group of answer choices Yes No

Transcript

00:01 We have these data for toronto and for ottawa, and we want to see if their level of satisfaction is basically the same, or, well, we want to compare the level of satisfaction.

00:14 And so it says in question a, which tests would you run, a pooled or a non -pooled? i would run a non -pooled because there isn't any indication that they would have their standard deviations it's unknown that their standard deviations are equal.

00:42 So we use a non -pooled.

00:44 Now on part b, it says that we're to use an alpha level of 0 .05, and if we decide to use a non -pooled test, we are told to use degrees of freedom of 60 .8.

01:02 So we'll use degrees of freedom of 60 .8.

01:07 And we want to have evidence whether the average satisfaction in toronto is, oh, part b says, do you have to assume that underlying conditions of satisfaction are normal? and no, because the sample sizes of the two are equal to 36 and the central limit theorem tells us that the sampling distribution would be approximately normal with those sample sizes.

01:40 So now part c says, based on your answer from a, we want to do a test to whether there's a higher average for toronto.

01:51 So we will be assuming that the mean for toronto is equal to that for ottawa, and alternately that the mean for toronto is higher than that for ontario.

02:03 Now it shows we have to compute the value of the test statistic and the t value, which will have 60 .8 degrees of freedom, is equal to the difference between their two means.

02:14 And the first mean is that 6 .3056 minus the mean of the other, which is 5 .7952, five, two, i believe, and then we'll take, divided by the standard deviation of the first, which is that 1 .6003, divided by that 36, and then we'll get that other standard deviation, which is that 2 .3963.

02:53 Now i'm actually gonna enter this into my software as a two -sample t test with statistics, and i'm gonna go back up here to look at that data again.

03:04 So 6 .3056, our sample standard deviation, 1 .6003, sample size is 36, and then that 5 .7952, 2 .3963, and 36.

03:19 And we're doing a greater than test, and we're gonna say no to pooling.

03:26 And we find out that this test statistic comes out to be a 1 .06, and it will round to three.

03:35 And the p -value for this test is equal to .1460.

03:41 And so at a 5 % significance level, we would fail to reject the null.

03:54 So there is insufficient evidence that the toronto mean is greater than the ottawa mean satisfaction score...